In article <email@example.com>, WM <firstname.lastname@example.org> wrote:
> On 29 Nov., 18:35, William Hughes <wpihug...@hotmail.com> wrote: > > On Nov 29, 8:05 am, WM <mueck...@rz.fh-augsburg.de> wrote: > > > > WM has conceded that you can use induction > > to show that every element of the list has > > a final 1, and that there is a constructive > > proof that the diagonal number does not have a > > final 1. > > > > WM has a new argument. > > > > > Use induction to show that the diagonal number cannot have more digits > > > than every entry of the list. > > > > This cannot be done. All you do is show that > > every one of an infinite number of different > > numbers, none of which is the diagonal number, > > cannot have more digits than every entry of the list. > > There is a simple proof by contradiction: > Assume that the diagonal (in the example-list > > 0.0 > 0.1 > 0.11 > 0.111 > ...) > > has a digit that is not in an entry of the list. This would mean that > the list has an end. That is wrong by definition. > Therefore every seqeunce of 1's in the diagonal is in an entry of the > list. > > Regards, WM
Your "sample list" must actually be 0.000... 0.1000... 0.11000... 0.111000... ... Whence an acceptable "anti-diagonal" is 1.111... which differs from every entry to the list.